now changing all undirected edges to directed there will still be no 2 students who are sending emails to each other because we changed undirected to directed so there are no cycles between vertices of length 2.

Now we got 5 remaining edges or emails to send as soon as we'll add these 5 directed edges, we'll get 6 more people at minimum.

Can you rephrase the third sentence ? It doesn’t make much sense.
– KanturaNov 8 at 19:33

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Do you mean "at least how many pairs of people have sent emails to each other"? Because if A and B exchange emails and A and C exchange emails, we cannot say that $3$ people have sent emails to each other as B and C have not done so.
– JensNov 8 at 22:45

Can you please rephrase the question? Do you mean "At least how many pairs of people sent emails to each other" or something like that?
– YiFanNov 9 at 8:40

I got the exact same question in a work book. Now i'm also confused how to interpret.
– Mk UtkarshNov 9 at 10:41

1 Answer
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Assuming the question is "What is the minimum number of pairs of students who sent emails to each other?", there must be at least $5$ pairs of students who sent emails to each. There are $\binom{10}{2}=45$ pairs of students, meaning that if we must distribute $10*5=50$ edges among $45$ pairs, then by the pigeonhole principle there must be at least $5$ pairs with $2$ emails. To finish the proof that this is the maximum lower bound, we can construct a graph with $50$ directed edges with exactly $5$ pairs with $2$ emails between them. This is not difficult, so I leave it to you.